Linear Algebra – Identify Mistake in Proof of Rank(A) + Rank(A-I) = n iff A² = A

idempotentslinear algebramatricesmatrix-rankprojection-matrices

Let $A$ be an $n\times n$ matrix over a field $F$. Show that $A^2=A$ if and only if $\text{rank}(A)+\text{rank}(A-I)=n$.

My attempt:

  1. $A^2=A \implies \text{rank}(A)+\text{rank}(A-I)=n$
    Using the lemma that $\text{rank}(AB)=\text{rank}(B)-\dim(\text{ker}(A)\bigcap \text{Im}(B))$, we can substitute $A$ for both $A$ and $B$ and get:
    $$\text{rank}(A^2)=\text{rank}(A)-\dim(\text{ker}(A)\bigcap \text{Im}(A))$$
    Since we have $A^2=A$, it follows that $\dim(\text{ker}(A)\bigcap \text{Im}(A))=0$, which implies that $\text{ker}(A)\bigcap \text{Im}(A)=\{0\}$. Now, if we suppose that $A^2x=0$, then the solution set of this equation is $\text{ker}(A)\bigcap \text{Im}(A)$. Hence, the only solution is the zero vector. By a theorem, this means that $A^2$ is invertible. Moreover, since $A^2=A$, we also have that $A$ is invertible. Therefore, we can multiply both sides by $A^{-1}$ and obtain $A=I$. This gives us $\text{rank}(A)=n$ and $\text{rank}(A-I)=0$. Thus, $\text{rank}(A)+\text{rank}(A-I)=n$ as required.

However, the conclusion $A=I$ is incorrect. There exist other matrices that satisfy $A^2=A$ but are not the identity matrix. See here for some examples. Where did I make a mistake in my proof?

Best Answer

Since we have $A^2=A$, it follows that $\dim(\text{ker}(A)\bigcap \text{Im}(A))=0$, which implies that $\text{ker}(A)\bigcap \text{Im}(A)=\{0\}$.

So far so good.

Now, if we suppose that $A^2x=0$, then the solution set of this equation is $\text{ker}(A)\bigcap \text{Im}(A)$.

This is the part that is incorrect. The solution set of this equation is the set of vectors $x$ satisfying $A^2 x = 0$, which will in this can be equal to $\ker(A)$ (since $\text{ker}(A)\bigcap \text{Im}(A)=\{0\}$). On the other hand, the set of vectors $Ax$ such that $A^2x = 0$ is indeed equal $\text{ker}(A)\bigcap \text{Im}(A)=\{0\}$. You seem to be confusing these distinct sets.

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